Question

Due to some force $F_1$, a body oscillates with period 4/5 second and due to other force $F_2$ it oscillates with the period of 3/5 sec. If both forces act simultaneously new period will be

• A. 0.72 s
• B. 0.64 s
• C. 0.48 s
• D. 0.36 s

Answer 1

Answer: (C)

0.48 s

for first case, ${\omega }_1^2=\frac{F_1}{mx}$

for second case, ${\omega }_2^2=\frac{F_2}{mx}$

When both forces applied, ${\omega }_3^2=\frac{F_1+F_2}{mx}$

$\Rightarrow {\omega }_3^2={\omega }_1^2+{\omega }_2^2$

$\Rightarrow {\omega }_3^2/4{\pi }^2={\omega }_1^2/4{\pi }^2+{\omega }_2^2/4{\pi }^2$

$\frac{1}{T_3^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2}$

$T_1=4/5=0.8s;T_2=3/5=0.6s$

$\Rightarrow T_3=12/25=0.48sec$